Problem: Divide the following complex numbers. $ \dfrac{-20+5i}{4-i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4+i}$ $ \dfrac{-20+5i}{4-i} = \dfrac{-20+5i}{4-i} \cdot \dfrac{{4+i}}{{4+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-20+5i) \cdot (4+i)} {(4-i) \cdot (4+i)} = \dfrac{(-20+5i) \cdot (4+i)} {4^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-20+5i) \cdot (4+i)} {(4)^2 - (-1i)^2} = $ $ \dfrac{(-20+5i) \cdot (4+i)} {16 + 1} = $ $ \dfrac{(-20+5i) \cdot (4+i)} {17} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20+5i}) \cdot ({4+i})} {17} = $ $ \dfrac{{-20} \cdot {4} + {5} \cdot {4 i} + {-20} \cdot {1 i} + {5} \cdot {1 i^2}} {17} $ Evaluate each product of two numbers. $ \dfrac{-80 + 20i - 20i + 5 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{-80 + 20i - 20i - 5} {17} = \dfrac{-85 + 0i} {17} = -5 $